Question

Last integral of the night for me\[\int e^{4x}\sqrt{1+e^{2x}}dx\]I just think it's fun.

Answer 1

hint:
\[\int\limits_{}^{}2 \cdot \frac{1}{2} e^{2x} e^{2x} \sqrt{1+e^{2x}} dx\]

Answer 2

\(e^{2x}=\tan^2(u)\) would work, I think.

Answer 3

without solving, my first instinct would be to do a u-sub for \[e^{2x}\]

Answer 4

omg why mr.math

Answer 5

don't make it hard

Answer 6

u=1+e^(2x)

Answer 7

\[u=1+e^{2x} => du= 2 e^{2x} dx ; u=1+e^{2x} => u-1=e^{2x}\]

Answer 8

\[\int\limits\limits_{}^{}2 \cdot \frac{1}{2} e^{2x} e^{2x} \sqrt{1+e^{2x}} dx \]
\[\frac{1}{2}\int\limits_{}^{} (u-1) \sqrt{u} du\]

Answer 9

I did it Mr.Math's way. Glad I asked, this seems much simpler.
G'nigh y'all :D

Answer 10

lol

Answer 11

boys are cute

Answer 12

;)

Answer 13

:)